The brakes applied to a car produce an acceleration of $6\, m s ^{-2}$ in the opposite direction of motion. If the car takes $2\, s$ to stop after the application of the brakes, calculate the distance it travels during this time.
The brakes applied to a car produce an acceleration of $6\, m s ^{-2}$ in the opposite direction of motion. If the car takes $2\, s$ to stop after the application of the brakes, calculate the distance it travels during this time.
Given $a=-6 m s ^{-2}, t=2 s , v=0 m s ^{-1}$
$(i)$ Using the equation $v=u+a t,$ we have
$u=v-a t=0-(-6) \times 2$
$=12 m s ^{-1}$
$(ii)$ Now, using the expression $S=u t+1 / 2 a t^{2}$
we have $S=12 \times 2+\frac{1}{2} \times-6 \times(2)^{2}=12 m$
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